Getting the wrong result for area between regions.

Question

This is what I'm given: $$y=cos(x)\;\;y={3x\over2\pi}\;\;with\;\;x\ge0$$

I asked earlier how to find the intersection of the two curves, and it is $x=0 \;and\;x={\pi\over3}$

So now I have $$\int_0^{\pi\over3}cos(x)-{3x\over2\pi}dx\;=\; sin(x)-{{3x^2}\over4\pi}$$

After putting the endpoints in: $$\left[{\sqrt3\over2}-{{{\pi\over3}^2}\over4\pi}\right]-0$$ And now... $${\sqrt3\over2}-{\pi\over12}$$ That is my final answer.

If I use an online calculator to check my answer it says it is not the same, which is true since when I whip the calculator out I get a different number than the online calculator.

Any help is appreciated. I could be wrong or the calculator could be wrong, I just want to check.

Thank you!

Answer 1

cos{0)=1, so the curves do not intersect at zero.

Answer 2

Graphs look like this. Try took into it